I need to know how precise a measurement system is, expressed as a standard deviation. I don't reall

Suppose you have ng=100 independent measurements from the distribution $\mathsf{N}\mathsf{o}\mathsf{r}\mathsf{m}(\mathrm{\mu <!--\; \mu \; -->},\mathrm{\sigma <!--\; \sigma \; -->}),$ where there are n=10 measurements on each of g=10 disjoint groups.
Let $S_i^2=\frac{1}{9}\underset{n=1}{\overset{10}{\sum <!--\; \sum \; -->}}(X_{ij}-<!--\; -\; -->{\stackrel{¯<!--\; ¯\; -->}{X}}_i{)}^2$ be the sample variance of group i out of g, where $X_{ij}$ is the $j$th measurement from group i and ${\stackrel{¯<!--\; ¯\; -->}{X}}_i=\frac{1}{10}\underset{j=1}{\overset{10}{\sum <!--\; \sum \; -->}}{X}_{ij}$ is the sample mean of the $i$th group.
Then the 'pooled' variance for the g groups is
$S_p^2=\frac{9S_1^2+9S_2^2+\cdots <!--\; \cdots \; -->+9S_{10}^2}{90}.$
$S_p^2$ is an unbiased estimate of the population variance ${\mathrm{\sigma <!--\; \sigma \; -->}}^2.$ For our normal data, its distribution is given by $90S_p^2/{\mathrm{\sigma <!--\; \sigma \; -->}}^2\sim <!--\; \sim \; -->\mathsf{C}\mathsf{h}\mathsf{i}\mathsf{s}\mathsf{q}(df=90).$
If necessary, this relationship would allow you to find a 95% confidence interval for ${\mathrm{\sigma <!--\; \sigma \; -->}}^2$ of the form $(90S_p^2/U,90S_p^2/L),$ where L and U cut probability 0.025 from the lower and upper tails of Chisq(df=90). For 90 'degrees of freedom' as in your example, L=65.65 and U=118.14 can be found from printed tables of the chi-squared distribution or by using software. (The result below is from R statistical software.)
qchisq(c(.025,.975), 90)## 65.64662 118.13589
Note: This kind of pooled variance is used in a one-factor analysis of variance (ANOVA) with $n$ replications on each of $g$ groups. It is called the 'denominator mean square' or 'mean square for error'. A one-factor ANOVA could be used to check whether it is reasonable to assume that all $g$ groups really do have the same population mean $\mathrm{\mu <!--\; \mu \; -->},$ as assumed in your example. You can find descriptions of such an ANOVA design in an intermediate level statistics book or online.